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,.

Sensor and Simulation Notes

Note 281,February 1983

EFFECT OF PULLING A WIRE OUT OF A CABLE BUNDLE ON THE :DISTRIBUTION OF CURRENT IN THE BUNDLE

. . ..”_-..-_ ... ---- .:.

Robert L. GardnerJames L. Gilbert

Mission Research Corporation

ABSTRACT

In this note, we apply a methodology of transmission line junctiontheory which considers the variation across a cable bundle in the con-tinuous variable approximation. The application is the calculation of thecurrent on a wire which has been pulled out. of a bundle to apply a currentprobe relative to the bundle current. For the 45-wire bundle we considerthere are about 20 surface wires, resulting 5 percent of the bundle currenton each wire for uniform excitation. The method predicts around 17 percent‘f the bundle current on the wire selected for measurement rather than 5Percent. This conclusion agrees with distributions of measured single wirecurrents in bundles which have minimums of 20 percent of the’bundle currenton each wire.

*

.. . ., ..

,= .— —

Sensor and Simulation Notes

Note 281

February 1983

EFFECT OF PULLING A WIRE OUT OF A CABLE BUNDLE ON THEDISTRIBUTION OF CURRENT IN THE BUNDLE

Robert L. GardnerJames L. Gilbert

Mission Research Corporation

ABSTRACT

In this note, we apply a methodology of transmission line junctiontheory which considers the variation across a cable bund!e in the con-tinuous variable approximation. The application is the calculation of thecurrent on a wire whtch has been pulled out of a bundle to apply a currentprobe relative to th~ bundle current. For the 45-wire bundle we considerthere are about 20 surface wires, resulting 5 percent of the bundle currenton each wire for uniform excitation. The method predicts around 17 percentof the bundle current on the wire selected for measurement rather than 5percent. This conclusion agrees with distributions of measured single wirecurrents in bundles which have minimums of 20 percent of the bundle currenton each wire.

transmission lines, theory, cables, wire, excitation

CONTENTS

Section Page

I DISCUSSION 4

II JUNCTION CONDITIONS 5

III APPLICATION TO SENSOR INSERTION 8

IV CONCLUSIONS 24

REFERENCES 25

APPENDIX A - POWER FI7ACTION SPLIT OFF ONTO WIRE 26

, ,

●

Figure

1

/ 2

3

4

5

6

7

8

Al

ILLUSTRATIO~iS

Geometry of a single wire pulled outwi res

of a bundle of

Current division as a function of frequency for thesimple 4-tube case

Image theory solution geometry

Current division for various frequencies as a function ofthe separation of the surfaces of the bundle and wire

Modification of geometry in Figure 1 showing additional3-wire section which allows first order treatment of thecapacitive (rather than conductive) coupling betweenthe wires in the bundle

Fraction of bundle current flowing on single wire pulledout from the bundle for the case of the wire and nearestneighbors treated as tubes separate from the main bundle

Geometry showing two wires pulled out of the bundle

Fraction of bundle current flowing on each of two wirespulled out from main bundle

Waveform used for the weighted average of the currentsplit off onto the small wire

Page

8

14

14

20

20

21

23

23

27

3

1. DISCUSSION

One of the results of various cable drive tests that have been run on

aircraft, missiles, and satellites is the d~stribution of currents induced

on wires in a bundle. In an example cable drive test treated here a bundle

of about 45 shield wires was studied in detail and individual shield cur-

rents were measured on each ot the 45 shields. As is usual for these kinds

of measurements the shield currents ‘were measured by pulling the wire out

of the bundle and enclosing the wire inside a current probe. A typical

distribution of individual shield peak currents measured was from 0.3 to

1.2 times the peak bundle current. Under uniform excitation and uniform

termination conditions each shield on the surface of the bundle should

carry 5 percent of the bundle current. There are, of course, phasing dif-

ferences for the individual incident currents in the real case, without

sensors. The same sort of distribution of peak currents also appears in

other systems with other excitation sources. In some aircraft measurements

a distribution of peak current 0.2~ IShield/IPeak ~ 2 was observed wqth

some individual wire currents observed as high as 3.5 times the bundle cur-

rent. The existance of a minimum peak current that is much larger than the

uniform excitation value indicates that there may be a perturbation of the

measurement due to the installation of the sensor. The cables were all

alike and terminated identically in a short. The individual cables were

coaxial cables with bra’ided shields and an exterior insulation layer.

As part of a program to develop methods of treating cable bundles with

large numbers of component wires, a methodology of treating the bundles

using a continuous variable approximat~on has been developed. The fol-

lowing is a subset of that treatment which incorporates the theory of junc-

tions (Ref. 1) in the continuous variable context. The current distribu-

tion for a wire pulled out of a bundle is then treated as an example.

Several particular geometries are treated to maximize the understanding

gleaned from the example problems.

4

. .

II. JUNCTION CONDITIONS

The above conditions determine the current flow in the continuous

representation of the cable bundle until a discontinuity is encountered.

Discontinuities, in this case, include branches, terminations, and, in some

sense, redistribution of wires within the bundle. The effects of each of

these discontinuities may be calculated using a continuous variable approx-

imation to the junction condition (Refs. 1 and 2) by computing the scat-

tering matrix of the junction.

Consider a three tube network in which a single bundle branches into

two smaller bundles according to a mapping operator, M. M operates only on

the coordinates and maps position rl in tube 1 to r2 in tube 2 or r3

in tube 3, depending on rl. There are some restrictions on the mapping.

First the area of tube 2 plus the area of tube 3 must equal that of tube 1

to conserve the total number of wires. In conserving area we do not con-

serve perimeter so that sufficient wires (unit areas) must be mapped from

the interior of tube 1 to fill out the perimeter of tubes 2 and 3. For

example, the splitting of circular tube 1 into two semi-circular tubes 2, 3

is performed by

(1)

The center wire goes to the edge of tube 2. From our solution to the prop-

agation equations above we have J(r,t) and ~(r,t) incident on the junction

from tube 1. We need the reflected waves into the three tubes, which may

be derived from Kirchhoff’s Laws for the junction. The charge density is

related to the potential in a wire through v2V = P/cO. The formal

● ●

●inverse of the capacitance matrix C (similar to .bbzcodv~ in operator form,

i.e. , CV = Q, where Cod is the capacitance between the wire of interest

and one o? its nearest neighbors, and b is one-half the wire spacjng; C and

the above differential operator are the same in that e~ther operating on V

yield Q. In one case V is a vector; in the other V is a continuous vari-

able, the infinite dimensional extension of the vector V.) is the inducta-

nce matr!x vZL. A formal inverse to the transverse Laplacian may be

found by considering a formal operator Lop which must have the prOpertY

Lopv$ = v (2)

The solution is

L~p=~dS’ G(r, r’) (3)

where G is the solution of

v~G(r,r’) = d(r-r’) (4)

and where S1 is the cross sect~on of the appropriate tube. The operator

approach is not necessary for those geometries that need numerical treat-

ment since the scattering matrix may be found knowing the self-capacitance

of each cell in the tube, without going through the operator step.

Kirchhoff’s Laws then require that the voltages for the simple connections

are the same at both ends of the connection and the current into the con-

nection must be balanced by a current flow out of the connection. Then

(5)

(6)

In this symbology V and I are functions of position and are continuous.

Let Y be an operator which converts V into J, i.e., J = YV. Note this

operator must act on the local voltage. That is, the voltage on tubes 2

6

* .

and 3, formed from NJ, uses the characteristic admittance operator for

those tubes. Since the characteristic admittance operator is proportional

to ~dS G, the Green’s function and integration surface for the correct tube

must be used.

the respective

formal problem

The Green’s function must obey the boundary conditions of

tube, requiring numerical solution of most problems. This

may now be solved.

In matrix-operator form, Kirchhoff’s Laws, for continuity of current

and voltage, are

(:,)(xl”) ‘0 (7)

where Z is some useful characteristic impedance and is included to provide

for a unitless matrix operator. The incident and reflected waves are

related by

“tot = Vine + Vref

~tot = ~inc - ~ref

for all tubes. Then

(z :) (~~c)+(:yz :Z) (:~f) ‘0 “0)

or in the scattering matrix-operator form

(i7:f)=-(:Yz :z)-’ (:Z :)(:2

(8)

(9)

(11)

Note that with proper selection of mapping function and characteristic

admittance matrix both redistribution of wires within a bundle and termina-

tions may be treated.

, .

t

111. APPLICATION TO SENSOR INSERTION

An immediate application of the above procedure is to determine the

effect of pulling a wire out of the bundle and placing a sensor around the

shield. Measurements have indicated that the phasing of the current on the

wires Is such that each individual shield carrjes a fifth to a third of the

bundle current, minimum. For the “phasing” to be such that all 45 wires in

the bundle to have large currents when measured, indicates there is some

perturbation by the sensor.

A single example of how current may be Increased Is to consider a large

bundle from which a single wire is split off, and all are terminated is a

short to ground. The appropriate geometry is shown in Figure 1. The sim-

ple circuit modeling this arrangement, for the case that the shields

(except that of the wire being pul led out) a~e assumed wel 1 connected

within a bundle, is a large inductor branching into small and large induc-

tors in parallel. For the example case there is an insulation layer, wh;ch

is treated in an approximate sense in a later section. The current wI1l

flow predominantly into the large inductor, i.e., the large wire. However>

since the inductance of the wire scales as a log and the division of the

current for the equal excitation case scales as the diameter of the bundle

more current will flow on the small wire than would have flowed jf the

Figure 1. Geometry of a sing” e wire pulled out of a bundle of wires.

8

separation had not occurred. The applicability of this example is yet to

be established, since the electrical connectivity between wires is not as

good as one might like. This connectivity problem will be considered later

in a 6-tube example.

Consider the breakout near the end of a cable bundle where one exterior

wire is removed from the bundle, and consider the first reflection of the

peak signal. If the wire is conductor 1 and the bundle is conductor 2, the

capacitance matrix is, approximately, before the breakout

and, after the breakout

C2 =

Cs o

\o

Cb

-cm

Cb + cm)

(12)

(13)

where Cb is the bundle over the ground plane, Cm is the mutual term

between the wires, Cs is the capacitance of a single wire above the

ground plane, and n is the number of perimeter wires. As a first approxi-

mation the mutual capacitance between the pulled out wire and the bundle

are assumed zero. A better calculation of those mutuals is given in the

next section. Consider frequencies low compared to the length over which

the section is removed.

Specific example values for the coefficients, for a 49 wire bundle, are

k =5cm

w= 108 s-1

dl = 1,4 cm

d2 = 0.2 cm

n = 22

9

+ .

●Cb = 4 x 10-11 F/m

C~ = 1.6 Y.10-11 F/m

Cm = 4 x 10-10 F/m (14)

and a typical bundle height is h = 2.4 cm. A useful figure 07 merit is

v/iil$= 60, where v is the propagation velocity of a signal along the trans-

mission line. The velocity is assumed constant through the bundle.

The characteristic admittance of the wire before the breakout is

Y1 = Vcl (16)

where v Is the propagation velocfity. The impedance of the breakout section

is

ZL = jw~L2 (17)

so the admittance is

~z

()

V2 Cs 0Y*=m c2=-

Jw2 ~ Cb

Using these sign conventions

Itrans>

linc

then the voltage and current must obey

Vi+vr=vt

Ii+ Ir=It

The voltage and currents are related by

(18)

(19)

10

Ii = YI Vi

Ir = -Yl Vr*

It = Y2 Vt

Solving these equations leads to

Vt = 2(Y1 + Y2) -1 Y1 vi

It = 2Y2(YI + Y2)-1 YI Vi

The useful Vi is

()‘ovi =

‘o

representing a uniform excitation.

Substituting the appropriate values of the adm’

above equations

()Cb/nIi=YIVi=vVo

Cb

and

where

/

VcbCb+cm+ c~ m

(Y1 + Y2)-1 =%P

\cm

(20)

(21)

(22)

ttance matrices into the

‘b + c “S

-ii- m ‘m )

(23)

(24)

2

I

2Cb Cbcm + v Cb

I

vzc~cbA =m+cbcm+y Cscb+ Cscm +~+ Cbcm +—~

m (jw)

11

●

F~nally, the transmitted current vector is

2VV0 vIt=r

()m

/c~c: vc;c~ \

+ Cmcbcs ~n I-I s + Cbcnlcs

( 2Cbcm

C3 vc~c:+#+cmc:+-

n J WE/

A couple of cases are of interest. If Cm + 0, (i.e, there is no

mutual coupling) and we keep the terms ordered

CbCb >> Cs >>y

then the transmitted current is

It ()Cbhl= 2“V0 Cb= ?Ii

or the transmitted current is twice the incident current, which is

pleasing.

In the ordering Cm >> Cb >> Cs >> Cb/n, the transmitted wave is (if

Cm >> vCs/ju~) for the tightly coupled bundle CaSe

()Csit = 2VV0

Cb

The measurements from which we take our numerjcal example ~s

Cb~cs>cm>cb>cs>r

(25)

(26)

(27)

(28)

(29)

12

,,

with

and the dominant terms turn out to be

()jw Cm

It = 2vV0 v

Cb

The fraction of current on the single wire is

(of, cm

—= 0.17V Cb

(30)

(31)

(32)

For a uniformly excited bundle each perimeter wire carries a current of

about 0.05 the bundle current so even this approximation suggests a factor

of 3.7 increase in the current on the wire.

Confirming this limiting estimate, a calculation of the current split

using Equation 21 was performed for frequencies between 1 MHz and 100 MHz.

The result is shown in Figure 2.

Effect of Mutual Terms on Current Redistribution

The expression for C2 in Equation 13 ignores the contribution of the

capacitance between the bundle and the wire pulled out. An estimate of the

mutual capacitance may be found by computing the elastance matrix for the

following geometry by image theory and inverting the matrix to find the

capacitance matrix. The geometry used for the image theory calculation is

shown in Figure 3. As shown, only six image charges are used in the cal-

culation.

13

10° ~ I I 1 [ I I I I I

10-1 z

10-$ /0 I I I I 1 I I I20 30 40 50 60 70 80 90 100

Frequency (MHZ)

Figure 2. Current division as a function of frequency for thesimple 4.tube case

t—-+-’-l

Ground Plane

t-a-il-ib

.

●

Figure 3. Image theory solution geometry.

14

If charges Q and Tare placed on the large and sma”

then the resulting potentials can be used to calculate

1 cylinders above,

the elastance matrix

elements. The charges ql and~l are located at the center of each of

the cylinders and are used to control the total charge on each of the

cylinders. q2 is the image of~l in the surface of the bundle. q3

is the image of ~2 + ~3 in the surface of the bundle. Similarly,I~2

and~3 are chosen to maintain a uniform potential on the surface “of the

wire. Note, the image theory series is truncated by assuming q2 + q3

generates only one image. A complete solution is an infinite series of

image charges, so this solution is best when the separation between bundle

and wire is more than a wire radius. A numerical solution is required for

better calculation of the capacitance matrix.’ Numerical treatment is

required since the wire actually enters the bundle when it gets close.

Further, the bundle is not a conductor but is a collectionof capacitively

coupled conductors.

The equations that the charges must obey are found by demanding that

the image plus impressed charge contributions result

around the cylinder surface (Ref. 3). The equations

q1+q2+q3=Q Ql+12+l’3=~

NI~2=.T

a(T2 + T3) b(q2 + q3)q3=- q3=-

R - b2/R R - a2/R

If we let

in a uniform potential

are:

(33)

(_Kl=ab $-1

R(R - a2/R) )

(34)

,

15

●

and

and

K* = ((l+ab 1 1

))R(R -bZ/R)- [R -a2/R)(R - b2/R) ,

b/R Q -’R2T1Tl=-

‘1 - ‘2

(1 1

‘1= ab

~- R(R -b*/R) )

K2 =

(

l+ab ( 1 1

R(R - a2/R) - (R - a2/R)(R - b2/R) ))

a/R v- K2Qql=-

1~

The other charges are found from

~z = -;T1

(35)

(37)

(38)

(39)

(40)

(41)

(Rz- a /R](RL- b /R)

16

(43)

Finally, the locations of the charges are also determined by the boundary

conditions (Ref. 3). For a coordinate system centered at the center of the

bundle and the x-axis through the center of the wire, the locations are:

‘1=0

‘2 = a2/R

‘3 = a2/(R - b2/R)

X4 = R - b/(R - a2/R)

‘5=R - b2/R

X6 =R (44)

where we have reindexed the charges to ql,2,3 ‘~1,2,3 and q4 ‘T3, q5 =

F*, and q6 =~l.

The elastance matrix elements are then found by adding up the contribu-

tions of the various line charges to the potential at the surface of the

bundle and wire.

The distances from the bundle surface for each of the charges are

‘l=a

d2 = a - a2/R

‘3=a- a2/(R - b2/R)

17

‘4=R-a - b2/[R - a2/R]

d5=R-a-b2/R

‘6= R-a

The distances from the wire surface

‘1 =R-b

‘2=R-b - a2/R

‘3=R -b- a2/(R - b2/R)

‘4=b - b2/(R - a2/R)

‘5=b - b2/R

‘6=b

The elastance matrix elements are then:

Q.Q Q=l

1

[

6

’21 =-—

2Tso z 1qikn ei - &n[2h+R]

-j=l

18

. *

& ●

(45)

(46)

. ...

Q=l Q-=IJ

[

61

’12 = -~ xqifm di - 8n[2h+R]

j=l

[

6

’22 = -~ zqi~n ei 1-ln[2h]

i=l

(47)

where the additional logarithmic terms account for reflection in the ground

plane.

The elastance matrix above was used in the calculation of C2 above as

a function of separation distance. Otherwise the problem was the same as

that described in the previous section. The current division as a function

of separation distance of the bundle and wire (surface to surface distance)

is shown in Figure 4. Near the bundle the pulled out wire arries about 5

percent of the current, as it should. Farther from the bundle, for 100 MHz

in particular, the fraction of current transferred to the wire becomes

larger, reaching 21 percent of the total current. Both self and mutual

capacitances are calculated with the image theory technique.

To confirm the conclusion of the effect of pulling out the wire two

more geometries were treated using the same junction theory technique. The

first of these geometries addresses the assumption, used above, that the

bundle i$ a solid conductor and is shown in Figure 5.

,.

This problem is now a junction of three conductors and the capacitance

matrix corresponding to those in Equation 12 is

I

(Cb/n + Cms -Cms o

2c@ + Cmb + Cmsc1 = - Cms-Cmb

)

(48)

o -Cmb (n-3)/n Cb + Cmb

19

100 : I I I 1 I I I

—100 MHz

-----10 MHz

●-.........1 MHz

__ —------ -------______ --------- --

n-mm . .......... ...........................**.....***.. .,.,...,.,,..,,.,,,,

-,~10 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0,010

Separation (m)

Figure 4. Current division for various frequencies as a function ofthe separation of the surfaces of the bundle and wire.

‘2

cms

cnlb

—

Figure 5. Modification of geometry in Figure 1 showing additional 3-wiresection which allows first order treatment of the capacitive(rather than conductive) coupling between the wires in thebundle.

20

.

●,.!

IA

where Cl represents the capacitance matrix be~ore the junction, and as

shown In Figure ~, ~s and Cmb are the mutual capacitances between the

intermediate layer and the wire and the bundle and intermediate layer

respectively. After the junction the capacitance between the wire and

.

intermediate layer is assumed negligible but the mutual capacitance between

the layer and the bundle is unchanged. The resulting capacitance matrix

after the junction is:

[

c~ (1 o

C*= o 2Cb/n + Cmb -Cmb

)

(49)

o -Cmb (n-S)/n c~ + cmb

Figure 6 shows the variation with frequency of the fraction of the total

current that is shifted from the wire. At frequencies associated with the

Frequency (MHz)

Figure 6. Fraction of bundle current flowing on single wire pulled outfrom the bundle for the case of the wire and nearest neighborstreated as tubes separate from the main bundle.

21

●

✎

+

rislltime of the electric drive pulses used here a substantial fraction of,.the total current is shifted to the wire with the sensor. The effect is

similar fo~ this geometry to that for the 2 x 2 (or 4 tube) problem.

A final geometry is shown in Figure 7 which has two sensor wire; pulled

out in the same way Troina bundle

case the initial capacitance

tance is Cm.

(Cb/n + Cm

Cl=*

-cm

matr

o

n

of 45 wit-es (22 surface wires). In this

ces where the bundle wire mutual capaci-

-Cm

~b/n + Cm -cm\

-cm Cb + 2Cm I

and

( )Cs o 0

C2= oCs o

0 0 Cb

(50)

(51)

The fraction of the total current on each of the two wires is shown in

Figure 8.

22

/. .

‘n l’=

I 2-/

Figure 7. Geometry showing two w res pulled out of the bund’e,

10°~ I I I I I

I

10-20 10 20 30I I I I I I I

~“ 90 100

Frequency (MHz)

Figure 8. Fractjon of bundle current flowing on each of two wirespulled out from the main bundle.

23

Iv. CONCLUSIUNS

I

It is evident from the previous calculations that the change in geome-

try caused by installing a sensor on a wire jn a bundle perturbs the dis-

tribution of current in the bundle, at least for the first peak. Our

calculations do not consider reflections. Certainly there are other

reasons for the distribution of peak currents being a larger ratio of the

bundle currents than one might expect intuitively. The particular phasing

of the components of the bundle is one of those reasons. However, these

calculations cast doubt on many of the minimum values for current peaks

observed on individual wires in bundles.

24

REFERENCES

1. Baum, C. E., T. K. Liu, and F. M. Tesche, On the Analysis of GeneralMulticonductor Transmission-Line Networks, Interaction Notes, Note 350,IMP Note Series, Air Force weapons Laboratory, Kirtland Air Force Base,Nrl.

2. Agrawal, A. K., H. M. Fowles, L. U. Scott, and S. H. Gurbaxani, TimeDomain Analysis of Multiconductor Transmission Lines with Branch-nn omogeneous Hedia, nteraction Notes,~ MP Note Series, Air

~atory, Kirtland Air Force tlas~, NM, August 1979.

3. Jackson, J. D., Classical Electrodynamics, 2nd Edition, John WileY,1975, p. 56.

25

APPENDIX A

POWER FRACTION SPLIT OFF ONTO WIRE

To confirm that the splitting of the current from tk bundle to the

single wire occurs for the excitation used in the sample measurements a

power spectrum weighted average was carried out. Cons~der the curr~nt on

the single wire as the product of a transfer function” T(w) and a bulk cur-

rent I(w). Then a power spectrum weighted average of the fraction F of the

bulk current on the single wire is given by:

(Al)

where R is a characteristic, constant resistance.

For the example cable drive test, a cable bundle was excited by driving

a meter long and a few centimeter wide plate parallel to the satellite

ground plane with a voltage across the plate and a termination impedance at

the center of the plate. The plate’s long dimension ran along the cable.

For this calculation the current was assumed proportional to dV/dt,

where V is the voltage applied to the parallel plate for the cable drive.

The waveform used for the applied electric field was developed from the

following curve fit:

(e-at e-yt

EP = 2.25 x 1051 + @~ - 1 + ~e-6t )

(A2)

26

a=2xlo9

B=Y=2X1O$3 ,

6= ZX 109

The resulting waveform is shown in Figure Al. This waveform was chosen to

eliminate the usual difficulty with the time derivates at t = O with the

double exponential. The result of this averaging process was that F = 16

percent of the bulk current was shifted to the wire pulled out of the bun-

dle, a factor of 3 more than one would expect for the uniform excitation

case.

1.6 X 105(

1.2 x 1051

0.8 X 105

0112

1

Time (ns)

Figure Al. Waveform used for the weighted average of the current splitoff onto the small wire.

27